Editing Ratio test (section)

====2. Raabe's test====

This extension is due to [[Joseph Ludwig Raabe]]. Define:

:<math>\rho_n \equiv n\left(\frac{a_n}{a_{n+1}}-1\right)</math>

(and some extra terms, see Ali, Blackburn, Feld, Duris (none), Duris2){{Clarify|date=September 2024}}

The series will:<ref name="Ali2008"/><ref name="Duris2009"/><ref name="Blackburn2012"/>
* Converge when there exists a ''c>''1 such that <math>\rho_n \ge c</math> for all ''n>N''.
* Diverge when <math>\rho_n \le 1</math> for all ''n>N''.
* Otherwise, the test is inconclusive.

For the limit version,<ref>{{mathworld|title=Raabe's Test|urlname=RaabesTest}}</ref> the series will:
* Converge if <math>\rho=\lim_{n\to\infty}\rho_n>1</math> (this includes the case ''ρ'' = ∞)
* Diverge if <math>\lim_{n\to\infty}\rho_n<1</math>.
* If ''ρ'' = 1, the test is inconclusive.

When the above limit does not exist, it may be possible to use limits superior and inferior.<ref name="Bromwich1908"/> The series will:
* Converge if <math>\liminf_{n \to \infty} \rho_n > 1</math>
* Diverge if <math>\limsup_{n \rightarrow \infty} \rho_n < 1</math>
* Otherwise, the test is inconclusive.

=====Proof of Raabe's test=====

Defining <math>\rho_n \equiv n\left(\frac{a_n}{a_{n+1}}-1\right)</math>, we need not assume the limit exists; if <math>\limsup\rho_n<1</math>, then <math>\sum a_n</math> diverges, while if <math>\liminf \rho_n>1</math> the sum converges.

The proof proceeds essentially by comparison with <math>\sum1/n^R</math>. Suppose first that <math>\limsup\rho_n<1</math>. Of course
if <math>\limsup\rho_n<0</math> then <math>a_{n+1}\ge a_n</math> for large <math>n</math>, so the sum diverges; assume then that <math>0\le\limsup\rho_n<1</math>. There exists <math>R<1</math> such that <math>\rho_n\le R</math> for all <math>n\ge N</math>, which is to say that <math>a_{n}/a_{n+1}\le \left(1+\frac Rn\right)\le e^{R/n}</math>. Thus <math>a_{n+1}\ge a_ne^{-R/n}</math>, which implies that
<math>a_{n+1}\ge a_Ne^{-R(1/N+\dots+1/n)}\ge ca_Ne^{-R\log(n)}=ca_N/n^R</math> for <math>n\ge N</math>; since <math>R<1</math> this shows that <math>\sum a_n</math> diverges.

The proof of the other half is entirely analogous, with most of the inequalities simply reversed. We need a preliminary inequality to use
in place of the simple <math>1+t<e^t</math> that was used above:  Fix <math>R</math> and <math>N</math>. Note that
<math>\log\left(1+\frac Rn\right)=\frac Rn+O\left(\frac 1{n^2}\right)</math>. So <math>\log\left(\left(1+\frac RN\right)\dots\left(1+\frac Rn\right)\right)
=R\left(\frac 1N+\dots+\frac 1n\right)+O(1)=R\log(n)+O(1)</math>; hence <math>\left(1+\frac RN\right)\dots\left(1+\frac Rn\right)\ge cn^R</math>.

Suppose now that <math>\liminf\rho_n>1</math>. Arguing as in the first paragraph, using the inequality established in the previous paragraph, we see that there exists <math>R>1</math> such that <math>a_{n+1}\le ca_Nn^{-R}</math> for <math>n\ge N</math>; since <math>R>1</math> this shows that <math>\sum a_n</math> converges.